You can define a distance in many ways, which one depends on what you are trying to do. The sum of the absolute differences is a simple one.
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copper.hatSep 19 '13 at 7:13

@copper.hat sum of absolute difference, I will think about that. But how? For example 1 : 3 : 5 and 3 : 4 : 5, one can say their absolute difference is 2 + 1 + 0 = 3 but if we take 1 : 3 : 5 and multiply it by 4/3, we get 4/3 : 4 : 20/3, and their difference will become 5/3 + 0 + 5/3 = 10 / 3, a bit larger than 3. So there will be cases where there are different difference for two proportions.
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Shane HsuSep 19 '13 at 7:43

1 Answer
1

This depends on which metric you define. Possible metrics would, for example, result in converting the $a:b:c$ to a vector $$\left( \begin{matrix}a\\b\\c\end{matrix} \right)$$
And using some standard $\mathbb R^3$ metrics, like
$$d(x,y) := \Vert x-y \Vert_p$$
For $1\leq p\leq \infty$.
The results vary with the choice of $p$. For $p=\infty$, for example we have with enumeration $y_1, y_2, y_3$ of your alternatives and $x$ the first:
$$\begin{align*}
d(x,y_1) & = 1 \\
d(x,y_2) & = 1 \\
d(x,y_3) & = 1
\end{align*}$$
so they are all "equally close".

However, any of these $p$ will yield a distance of $1$, since the changes are always to add one to one "component".

Another choice for converting $a:b:c$ to a vector containing "proportions", would be
$$\vec{x} = \left(\begin{matrix}a/b\\b/c\\c/a\end{matrix}\right)$$

So the idea is to turn my proportion into vector for calculation. And use Uniform Norm?
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Shane HsuSep 19 '13 at 15:27

Well, essentially you can use any $\mathbb R^3$ norm you are fond of. $p=\infty$ or $p=2$ should yield the most intuitive results. I recommend the euclidean norm ($p=2$) because it "punishes" all deviations and not only the highest.
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AlexRSep 19 '13 at 20:04

@ShaneHsu I've added MATLAB code for my recommended example and given the output. That seems an "intuitive" result to me...
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AlexRSep 19 '13 at 20:14